Add $ \dfrac{x^2}{x-5} $ and $ \dfrac{25}{5-x} $ to get $ \dfrac{ \color{purple}{ -x^2+25 } }{ -x+5 }$.

To add raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $\color{blue}{ -1 }$.

$$ \begin{aligned} \frac{x^2}{x-5} + \frac{25}{5-x} & = \frac{ x^2 \cdot \color{blue}{ \left( -1 \right) }}{ \left( x-5 \right) \cdot \color{blue}{ \left( -1 \right) }} + \frac{ 25 }{ 5-x } = \\[1ex] &=\frac{ \color{purple}{ -x^2 } }{ -x+5 } + \frac{ \color{purple}{ 25 } }{ -x+5 }=\frac{ \color{purple}{ -x^2+25 } }{ -x+5 } \end{aligned} $$

Simplify $ \dfrac{-x^2+25}{-x+5} $ to $ x+5$.

Factor both the denominator and the numerator, then cancel the common factor. $\color{blue}{x-5}$.

$$ \begin{aligned} \frac{-x^2+25}{-x+5} & =\frac{ \left( -x-5 \right) \cdot \color{blue}{ \left( x-5 \right) }}{ \left( -1 \right) \cdot \color{blue}{ \left( x-5 \right) }} = \\[1ex] &= \frac{x+5}{1} =x+5 \end{aligned} $$